Pointer behavior in functions in Go - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to an integer variable.

Go

var x int = 10
var p [1] = &x

Drag options to blanks, or click blank then click option'

Aint

B*int

C&int

Dint*

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of '*int' for pointer type.

Using '&int' which is not a valid type.

Using 'int*' which is not Go syntax.

In Go, a pointer to an int is declared as *int. The & operator gets the address of x.

2fill in blank

medium

Complete the code to modify the value of an integer through a pointer inside a function.

Go

func changeValue(n [1]) {
    *n = 20
}

func main() {
    x := 10
    changeValue(&x)
    fmt.Println(x) // Output: 20
}

Drag options to blanks, or click blank then click option'

A*int

Bint

Cint*

D&int

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of '*int' so the function modifies a copy.

Using invalid pointer syntax like 'int*' or '&int'.

The function parameter must be a pointer to int (*int) to modify the original variable's value.

3fill in blank

hard

Fill both blanks to make the increment function modify x correctly.

Go

func increment(n [1]) {
    (*n)++
}

func main() {
    x := 5
    increment([2]x)
    fmt.Println(x) // Output: 6
}

Drag options to blanks, or click blank then click option'

Aint

B&

C*int

Dpointer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of '*int' in the function parameter.

Passing 'x' instead of '&x' to the function.

The function parameter must be a pointer to int (*int), and the argument must be the address of x using &x.

4fill in blank

hard

Fill both blanks to create a function that swaps two integers using pointers.

Go

func swap(a, b [1]) {
    temp := *a
    *a = *b
    *b = temp
}

func main() {
    x, y := 1, 2
    swap([2]x, &y)
    fmt.Println(x, y) // Output: 2 1
}

Drag options to blanks, or click blank then click option'

A*int

B&

Cint

Dpointer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of '*int' for parameters.

Passing variables directly instead of their addresses.

The function parameters must be pointers to int (*int), and the arguments must be addresses of variables using &.

5fill in blank

hard

Fill all three blanks to create a function that doubles the value of an integer using a pointer and returns the new value.

Go

func doubleValue(n [1]) [2] {
    *n = *n * 2
    return [3]
}

func main() {
    x := 7
    result := doubleValue(&x)
    fmt.Println(result, x) // Output: 14 14
}

Drag options to blanks, or click blank then click option'

A*int

Bint

C*n

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' instead of '*int' for the parameter.

Returning the pointer variable instead of the dereferenced value.

Using wrong return type.

The function parameter is a pointer to int (*int), the return type is int, and the function returns the dereferenced value *n.

Practice

1. What does passing a pointer to a function in Go allow you to do?

easy

A. Automatically convert the variable to a string

B. Create a copy of the variable that cannot be changed

C. Modify the original variable's value inside the function

D. Prevent the function from accessing the variable

Solution

1. Step 1: Understand pointer passing

   Passing a pointer sends the address of the variable, not a copy.

2. Step 2: Effect on original variable

   Inside the function, using the pointer lets you change the original variable's value.

3. Final Answer:

   Modify the original variable's value inside the function -> Option C

4. Quick Check:

   Pointer lets function change original variable [OK]

Hint: Pointers let functions change original variables directly [OK]

Common Mistakes:

• Thinking pointers create copies instead of referencing original
• Confusing pointer with value passing
• Assuming pointers prevent modification

2. Which of the following is the correct way to declare a function that takes a pointer to an int in Go?

easy

A. func updateValue(val *int) {}

B. func updateValue(val int) {}

C. func updateValue(*val int) {}

D. func updateValue(&val int) {}

Solution

1. Step 1: Recall pointer syntax in function parameters

   In Go, a pointer parameter is declared with * before the type, like *int.

2. Step 2: Check each option

   func updateValue(val *int) {} correctly uses *int as the parameter type.

3. Final Answer:

   func updateValue(val *int) {} -> Option A

4. Quick Check:

   Pointer parameter syntax = *type [OK]

Hint: Pointer parameters use * before type, not variable name [OK]

Common Mistakes:

• Placing * before variable name instead of type
• Using & in parameter declaration
• Omitting * for pointer type

3. What will be the output of this Go program?

package main
import "fmt"
func change(val *int) {
    *val = 10
}
func main() {
    x := 5
    change(&x)
    fmt.Println(x)
}

medium

A. 5

B. 10

C. 0

D. Compilation error

Solution

1. Step 1: Understand pointer usage in change function

   The function receives a pointer to x and sets the value at that address to 10.

2. Step 2: Trace main function execution

   x starts as 5, change(&x) updates x to 10, then fmt.Println prints x.

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Pointer lets function update x to 10 [OK]

Hint: Pointer lets function change original variable value [OK]

Common Mistakes:

• Expecting original value 5 to print
• Confusing pointer dereference syntax
• Thinking function works without & operator

4. Identify the error in this Go code snippet:

func update(val *int) {
    val = 20
}

func main() {
    x := 5
    update(&x)
    fmt.Println(x)
}

medium

A. Cannot assign int to pointer variable directly

B. Missing & operator when calling update

C. Function update should return int

D. x is not initialized

Solution

1. Step 1: Analyze assignment inside update function

   val is a pointer to int, but val = 20 tries to assign int to pointer variable, which is invalid.

2. Step 2: Correct way to change value via pointer

   Use *val = 20 to assign 20 to the value pointed by val.

3. Final Answer:

   Cannot assign int to pointer variable directly -> Option A

4. Quick Check:

   Assign to *val, not val [OK]

Hint: Assign to *pointer, not pointer variable itself [OK]

Common Mistakes:

• Assigning value directly to pointer variable
• Forgetting to dereference pointer
• Confusing pointer and value types

5. Given this Go code, what will be the output?

package main
import "fmt"
func swap(a, b *int) {
    *a, *b = *b, *a
}
func main() {
    x, y := 1, 2
    swap(&x, &y)
    fmt.Println(x, y)
}

hard

A. Compilation error

B. 1 2

C. 0 0

D. 2 1

Solution

1. Step 1: Understand swap function behavior

   swap takes pointers to two ints and swaps their values using pointer dereferencing.

2. Step 2: Trace main function execution

   x=1, y=2 initially; swap(&x, &y) swaps values, so x becomes 2 and y becomes 1.

3. Final Answer:

   2 1 -> Option D

4. Quick Check:

   Pointer swap exchanges values correctly [OK]

Hint: Swap values by dereferencing pointers [OK]

Common Mistakes:

• Expecting original order after swap
• Confusing pointer addresses with values
• Not using * to dereference pointers